Question

For the reaction system $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ which has already reached a state of equilibrium, predict the effect that each of the following changes will have on the position of the equilibrium. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. The pressure of hydrogen is increased by injecting an additional mole of hydrogen gas into the reaction vessel. b. Carbon monoxide gas is removed as it forms by use of a chemical absorbent or "scrubber." c. An additional amount of solid carbon is added to the reaction vessel.

Short answer

For the given reaction system at equilibrium: a. Increasing the pressure of hydrogen gas will cause the equilibrium to shift to the left, where there is fewer moles of gas. b. Removing carbon monoxide gas will cause the equilibrium to shift to the right, where carbon monoxide is being produced. c. Adding solid carbon will not affect the position of the equilibrium, as it doesn't change the concentration, pressure, or temperature of the system.

Step-by-step solution

Part (a): Increasing pressure of hydrogen gas

For the first change, we increase the pressure of hydrogen by injecting an additional mole of hydrogen gas into the reaction vessel. According to Le Chatelier's principle, the system will adjust to counteract the increase in pressure. In this case, the reaction will shift to the side with fewer moles of gas to decrease the pressure. In the given reaction system: \[\mathrm{C}(s)+\mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}(g)\] There are 2 moles of gas on the right side and 1 mole of gas on the left side. Therefore, to decrease the pressure, the reaction will shift to the left side.

Part (b): Removing carbon monoxide gas

The second change involves the removal of carbon monoxide gas as it forms by using a chemical absorbent or "scrubber." According to Le Chatelier's principle, the system will adjust to counteract the removal of carbon monoxide gas. In this case, the reaction will shift to the side where carbon monoxide is being produced to compensate for the loss. Therefore, the equilibrium will shift to the right side.

Part (c): Adding solid carbon to the reaction vessel

The third change involves adding an additional amount of solid carbon to the reaction vessel. In this reaction system, carbon is in solid form, so it doesn't affect the concentration or pressure of gases in the reaction. According to Le Chatelier's principle, only changes in concentration, pressure, and temperature can affect the position of equilibrium in a reaction system. Since adding solid carbon doesn't affect any of these factors, the equilibrium position will not be affected by the addition of solid carbon.

Key concepts

Equilibrium Shift

Le Chatelier's Principle offers a straightforward way to predict how the equilibrium position of a chemical reaction will respond to various changes. This principle states that if a system in equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust to partially counteract the effect of the change and establish a new equilibrium. In our example reaction \(\mathrm{C}(s) + \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{CO}(g)\), different scenarios prompt different equilibrium shifts.
- **Increase in Hydrogen's Pressure**: Adding more hydrogen gas increases pressure. The system tries to lower this pressure by shifting to the direction with fewer moles of gas. Since there is 1 mole of gas on the left side and 2 on the right, the shift will be to the left.- **Removal of Carbon Monoxide**: As CO gas is removed, the system compensates by shifting to the right to produce more CO.
Understanding how equilibrium shifts help us control chemical reactions and predict their outcomes under changing conditions.

Reaction Pressure

Pressure changes affect the equilibrium of reactions that involve gases. Le Chatelier's principle helps us understand that the system will respond in a way that opposes the change. When the pressure in a system is increased by adding more gas, the equilibrium tends to shift in the direction that produces fewer gas molecules. This is important when considering industrial applications where optimizing yield is crucial.

For the reaction of interest, increasing the pressure, such as by adding hydrogen gas, results in a shift to the left. This is because the left side has a lower total number of gaseous molecules (1 versus the 2 on the right). Pressure impacts are less noticeable in reactions where solids are a reactant or product because solids do not contribute to the system's overall pressure. Instead, gaseous components are the main focus when discussing pressure effects on equilibrium.

Solid Reactants

In chemical equilibrium studies, solid reactants are often viewed differently than gases or aqueous solutions. Their presence or concentration does not directly affect the position of equilibrium. This is because solids have a defined density and volume which means they cannot change concentration like gases or solutions can. For the reaction \(\mathrm{C}(s) + \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{CO}(g)\), carbon is a solid reactant.
- **Additions of Solids**: Adding more of a solid reactant like carbon will not shift the position of equilibrium because solids don’t affect the gaseous equilibrium pressures or concentrations.- **Why Solids Don’t Shift Equilibrium**: They do not appear in equilibrium constant expressions, which only involve concentrations (for gases, partial pressures). Therefore, the equilibrium constant remains unaffected by solid reactants.
While solid reactants are crucial for the reaction process, their addition or removal generally doesn’t change the equilibrium for reactions involving gases.